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CIE 9709 2023 November Paper 63 Q5

A Level / CIE / S2

CIE 9709 2023 Nov Paper 63 Paper · Question 5

题目

Problem

In the past the number of enquiries per minute at a customer service desk has been modelled by a random variable with distribution Po(0.31)\mathrm{Po}(0.31). Following a change in the position of the desk, it is expected that the mean number of enquiries per minute will increase. In order to test whether this is the case, the total number of enquiries during a randomly chosen 55-minute period is noted. You should assume that a Poisson model is still appropriate.

Given that the total number of enquiries is 55, carry out the test at the 2.5%2.5\% significance level.

(5)
题目中文翻译

过去，某客服台每分钟收到的咨询次数用服从 Po(0.31)\mathrm{Po}(0.31) 分布的随机变量描述。客服台搬到新位置后，预计每分钟的平均咨询次数会上升。为了检验是否如此，随机选取一段 55 分钟的时间，记录咨询总次数。假设泊松模型仍然适用。

已知咨询总次数为 55，在 2.5%2.5\% 的显著性水平下进行检验。

(5)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

解法一

思路

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以每分钟的平均询问数为参数建立右尾检验。原假设下五分钟总数参数为1.55，计算至少5次的概率。

答题过程

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Let λ\lambda be the population mean number of enquiries per minute after the change.

H0:λ=0.31,H1:λ>0.31.H_0:\lambda=0.31,\qquad H_1:\lambda>0.31.

Under H0H_0, the five-minute total Q∼Po(5(0.31))=Po(1.55)Q\sim\mathrm{Po}(5(0.31))=\mathrm{Po}(1.55). For the observed total 5,

 P(Q≥5∣H0)= 1−e−1.55(1+1.55) −e−1.55(1.5522!+1.5533!) −e−1.551.5544!= 0.0210283….\begin{align*} &\,\mathrm{P}(Q\geq5\mid H_0)\\[4mm] =&\,1-e^{-1.55}(1+1.55)\\[4mm] &\,-e^{-1.55}\bigg(\frac{1.55^2}{2!}+\frac{1.55^3}{3!}\bigg)\\[4mm] &\,\hspace{2pt}-e^{-1.55}\frac{1.55^4}{4!}\\[4mm] =&\,0.0210283\ldots. \end{align*}

Since 0.0210283…<0.0250.0210283\ldots<0.025, we reject H0H_0. There is sufficient evidence at the 2.5% significance level that the mean number of enquiries per minute has increased.